Remove all lines in file older than 24 hours - bash

Question

Ive seen a lot of questions regarding removing files that are older than x number of hours. I have not seen any pertaining to removing lines in a file older than x number of hours.
Here is an example of the log I am dealing with. For the sake of the example, assume current time is 2016-12-06 06:08:48,594
2016-12-05 00:44:48,194 INFO this line should be deleted
2016-12-05 01:02:10,220 INFO this line should be deleted
2016-12-05 05:59:10,540 INFO this line should be deleted
2016-12-05 06:08:10,220 INFO this line should be deleted
2016-12-05 16:05:30,521 INFO do not delete this line
2016-12-05 22:23:08,623 INFO do not delete this line
2016-12-06 01:06:28,323 INFO do not delete this line
2016-12-06 05:49:55,619 INFO do not delete this line
2016-12-06 06:06:55,419 INFO do not delete this line
I realize that it might be easier to do this in python or Perl but this needs to be done in bash. That being said, please post any and all relevant answers.
So far Ive tried using sed, awk, etc to convert the timestamps to seconds.
#! /bin/bash
TODAY=$(date +%Y-%m-%d)
# one day ago
YESTERDAY=$(date -d #$(( $(date +"%s") - 86400)) +%Y-%m-%d)
REPORT_LOG=report_log-$TODAY.log
# current date in seconds
NOW=$(date +%s)
# oldest date in the log trimmed by timestamp
OLDEST_DATE=$(head -1 $REPORT_LOG | awk '{print $1" "$2}')
# oldest date converted to seconds
CONVERT_OLDEST_DATE=$(date -d "$OLDEST_DATE" +%s)
TIME_DIFF=$(($NOW-$CONVERT_OLDEST_DATE))
# if difference is less than 24 hours, then...
if [ $TIME_DIFF -ge 86400 ]; then
LATEST_LOG_TIME=$(tail -1 $REPORT_LOG | awk '{print $2}'| cut -c 1-8)
RESULTS=$(awk "/${YESTERDAY} ${LATEST_LOG_TIME}/{i++}i" $REPORT_LOG)
if [ -z $RESULTS]; then
awk "/${YESTERDAY} ${LATEST_LOG_TIME}/{i++}i" $REPORT_LOG > $REPORT_LOG.tmp && mv $REPORT_LOG.tmp $REPORT_LOG
else
echo "Out of ideas at this point"
fi
else
echo "All times newer than date"
fi
The problem with my above snippet is that it relies on a date to repeat itself for the awk to work, which is not always the case. There are hour long gaps in the log files so it is possible for the last line's date (ex. 2016-12-06 06:06:55) to be the only time that date appears. If the timestamp has not previously appeared, my script will delete all results before the matched timestamp.
Any and all help is appreciated.

Answer 1

awk to the rescue!
$ awk -v d="2016-12-05 06:08:48,594" '($1 " " $2) > d' file
will print the newer entries. Obviously, you want to create the date dynamically.
Ignoring the milliseconds part to simplify, you can use
$ awk -v d="$(date --date="yesterday" "+%Y-%m-%d %H:%m:%S,999")" ...
Note that lexical comparison works only for your hierarchial formatted date (why don't everybody use this?), for any other format you are better off converting to seconds from epoch and do numerical comparison on integers

Answer 2

Do the dates in times since the Unix epoch, using the format string +%s. For instance:
yesterday=$(date --date="yesterday" +%s)
Then interpret the dates you've extracted with awk or similar like:
dateInUnixEpoch=$(date --date="$whateverDate" +%s)
Then just compare the dates:
if [ "$yesterday" -ge "$dateInUnixEpoch" ];
then do whatever to delete the lines
fi